(x^2+2x-7)-(2x+1)=0

Solution for (x^2+2x-7)-(2x+1)=0 equation:

(x^2+2x-7)-(2x+1)=0

We get rid of parentheses

x^2+2x-2x-7-1=0

We add all the numbers together, and all the variables

x^2-8=0

a = 1; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·1·(-8)
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*1}=\frac{0-4\sqrt{2}}{2}
=-\frac{4\sqrt{2}}{2}
=-2\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*1}=\frac{0+4\sqrt{2}}{2}
=\frac{4\sqrt{2}}{2}
=2\sqrt{2}
$